I came up with the following seemingly true statement:
$$\int_0^\infty \frac{e^{ix}}{x^{\alpha}} \, dx $$
for $\alpha \in (0,1)$ exists as an improper integral. Also, is it possible to compute the integral? For instance, the value of
$$ \int_0^\infty \frac{e^{ix}}{ \sqrt{x}} \, dx = ? $$
My thoughts on convergence:
The improper integral $\int_0^N \frac{e^{ix}}{x^{\alpha}} \, dx $ exists by DCT for all $N \in \mathbb{N}$. So it suffices to consider the cases, $$\int_N^\infty \frac{\cos x}{x^{\alpha}} \, dx , \int_N^\infty \frac{\sin x }{x^{\alpha} } \, dx .$$
Using integration by parts, we have
$$ \int_N^M \frac{ \cos x }{x^{\alpha} } = \Big[ \frac{ \sin x }{x^{\alpha} } \Big]_N^M + \alpha \int_N^M \frac{ \sin x }{ x^{\alpha+1} } \, dx $$
which converges as $M \rightarrow \infty$. Since the latter is dominated by $\int_N^M \frac{1}{x^{\alpha+1}} \, dx $ of which converges. A similar argument holds for imaginary part.
Is this argument correct?
Best Answer
Your argument is correct.
Furthermore, it is not too difficult to compute the value of your integral using some contour integration.
Cauchy's Integral Theorem supports the change of variables $x\mapsto ix$, which changes the path of integration from $[0,R]$ to $[0,-iR]$, followed by the change of path using the closed contour $$ [0,R]\cup Re^{-i[0,\pi/2]}\cup[-iR,0] $$ which contains no singularities of the integrand, giving $$ \begin{align} \int_0^\infty x^{-\alpha}e^{ix}\,\mathrm{d}x &=e^{(1-\alpha)\pi i/2}\int_0^\infty x^{-\alpha}e^{-x}\,\mathrm{d}x\\ &=\bbox[5px,border:2px solid #C0A000]{e^{(1-\alpha)\pi i/2}\Gamma(1-\alpha)} \end{align} $$ Since $\Gamma\left(\frac12\right)=\sqrt\pi$, the value of the integral for $\alpha=\frac12$ is $$ \int_0^\infty x^{-1/2}e^{ix}\,\mathrm{d}x=\sqrt{\frac\pi2}\,(1+i) $$